Answer
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Hint: In elementary algebra, root rationalisation is a process by which radicals in the denominator or numerator of an algebraic fraction are eliminated. To rationalize a term we need to multiply it with a term such that the square root, cube root etc is eliminated. We multiply with the same term in denominator and numerator to maintain the original value of the fraction which is to be rationalized.
Complete step by step answer:
Here we have the fraction \[\sqrt{\dfrac{\cos x}{\sin x}}\] where we need to rationalize the numerator.
In fraction we have two term numerator and denominator. The term above the line is called numerator and the term below the line is called denominator. For example in the fraction
\[\dfrac{2}{3}\], the above term that is 2 is numerator and the below term, 3, is the denominator.
Thus we can clearly see in the term \[\sqrt{\dfrac{\cos x}{\sin x}}\] the numerator is \[\sqrt{\cos x}\] and denominator is \[\sqrt{\sin x}\]
Now to rationalize the numerator we need to multiply it with a term such that the square root is removed. As we know multiplying two same square root terms result in same term , we will multiply both denominator and numerator by \[\sqrt{\cos x}\]
\[\Rightarrow \sqrt{\dfrac{\cos x}{\sin x}}\left( \dfrac{\sqrt{\cos x}}{\sqrt{\cos x}} \right)\]
(We know that \[\dfrac{\sqrt{\cos x}}{\sqrt{\cos x}}=1\] and multiplying anything with 1 will not alter its value)
\[\Rightarrow \sqrt{\dfrac{\cos x(\cos x)}{\sin x(\cos x)}}\]
\[\Rightarrow \sqrt{\dfrac{{{(\cos x)}^{2}}}{\sin x(\cos x)}}\]
\[\Rightarrow \dfrac{\sqrt{{{(\cos x)}^{2}}}}{\sqrt{\sin x(\cos x)}}\] , (by using the property \[\Rightarrow \sqrt{\dfrac{a}{b}}\Leftrightarrow \dfrac{\sqrt{a}}{\sqrt{b}}\] )
\[\Rightarrow \dfrac{(\cos x)}{\sqrt{\sin x\cos x}}\] , (by using the property \[\Rightarrow \sqrt{{{a}^{2}}}=a\] )
Hence the rationalized term required is \[\dfrac{(\cos x)}{\sqrt{\sin x\cos x}}\] , (only numerator rationalized).
Note: The rationalization of a fraction or a term is done to simplify it and to find out the actual value of the term in a more simple form. Generally the rationalization is done for the denominator, unlike in the above question, to make the denominator the same/rational in case there are multiple terms in an expression.
Complete step by step answer:
Here we have the fraction \[\sqrt{\dfrac{\cos x}{\sin x}}\] where we need to rationalize the numerator.
In fraction we have two term numerator and denominator. The term above the line is called numerator and the term below the line is called denominator. For example in the fraction
\[\dfrac{2}{3}\], the above term that is 2 is numerator and the below term, 3, is the denominator.
Thus we can clearly see in the term \[\sqrt{\dfrac{\cos x}{\sin x}}\] the numerator is \[\sqrt{\cos x}\] and denominator is \[\sqrt{\sin x}\]
Now to rationalize the numerator we need to multiply it with a term such that the square root is removed. As we know multiplying two same square root terms result in same term , we will multiply both denominator and numerator by \[\sqrt{\cos x}\]
\[\Rightarrow \sqrt{\dfrac{\cos x}{\sin x}}\left( \dfrac{\sqrt{\cos x}}{\sqrt{\cos x}} \right)\]
(We know that \[\dfrac{\sqrt{\cos x}}{\sqrt{\cos x}}=1\] and multiplying anything with 1 will not alter its value)
\[\Rightarrow \sqrt{\dfrac{\cos x(\cos x)}{\sin x(\cos x)}}\]
\[\Rightarrow \sqrt{\dfrac{{{(\cos x)}^{2}}}{\sin x(\cos x)}}\]
\[\Rightarrow \dfrac{\sqrt{{{(\cos x)}^{2}}}}{\sqrt{\sin x(\cos x)}}\] , (by using the property \[\Rightarrow \sqrt{\dfrac{a}{b}}\Leftrightarrow \dfrac{\sqrt{a}}{\sqrt{b}}\] )
\[\Rightarrow \dfrac{(\cos x)}{\sqrt{\sin x\cos x}}\] , (by using the property \[\Rightarrow \sqrt{{{a}^{2}}}=a\] )
Hence the rationalized term required is \[\dfrac{(\cos x)}{\sqrt{\sin x\cos x}}\] , (only numerator rationalized).
Note: The rationalization of a fraction or a term is done to simplify it and to find out the actual value of the term in a more simple form. Generally the rationalization is done for the denominator, unlike in the above question, to make the denominator the same/rational in case there are multiple terms in an expression.
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